Question: Find the limit as $x$ approaches positive infinity. $\lim_{x\to\infty}\dfrac{3x-1}{\sqrt{x^2-6}}=$
Solution: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x$, let's divide by $x$. In the denominator, let's divide by $\sqrt{x^2}$, since for positive values, $x=\sqrt{x^2}$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{3x-1}{\sqrt{x^2-6}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{3x-1}{x}}{\dfrac{\sqrt{x^2-6}}{\sqrt{x^2}}} \gray{\text{Divide sides by }x=\sqrt{x^2}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to\infty}\dfrac{\dfrac{3\cancel{x}}{\cancel{x}}-\dfrac{1}{x}}{\sqrt{\dfrac{1\cancel{x^2}}{\cancel{x^2}}-\dfrac{6}{ x^2}}} \\\\ &=\lim_{x\to\infty}\dfrac{3-\dfrac{1}{x}}{\sqrt{1-\dfrac{6}{x^2}}} \\\\ &=\lim_{x\to\infty}\dfrac{3-0}{\sqrt{1-0}} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{3}{\sqrt{1}} \\\\ &=\dfrac{3}{1} \\\\ &=3 \end{aligned}$ In conclusion, $\lim_{x\to\infty}\dfrac{3x-1}{\sqrt{x^2-6}}=3$.